/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A 220 -kg speedboat is negotiati... [FREE SOLUTION] | 91Ó°ÊÓ

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A 220 -kg speedboat is negotiating a circular turn (radius \(=32 \mathrm{m})\) around a buoy. During the turn, the engine causes a net tangential force of magnitude \(550 \mathrm{N}\) to be applied to the boat. The initial tangential speed of the boat going into the turn is \(5.0 \mathrm{m} / \mathrm{s}\). (a) Find the tangential acceleration. (b) After the boat is \(2.0 \mathrm{s}\) into the turn, find the centripetal acceleration.

Short Answer

Expert verified
(a) Tangential acceleration is 2.5 m/s². (b) Centripetal acceleration after 2 seconds is 3.125 m/s².

Step by step solution

01

Calculate Tangential Acceleration

To find the tangential acceleration, use Newton's second law of motion. According to it, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Given the net tangential force is 550 N and the mass of the speedboat is 220 kg, the tangential acceleration \( a_t \) is given by \( a_t = \frac{F_t}{m} \), where \( F_t = 550 \, N \) is the tangential force. Plug in the values and solve: \( a_t = \frac{550 \, \text{N}}{220 \, \text{kg}} = 2.5 \, \text{m/s}^2 \).
02

Find Tangential Speed After 2 Seconds

The tangential speed after 2 seconds can be found using the formula for velocity under constant acceleration: \( v = v_0 + a_t t \), where \( v_0 = 5.0 \, \text{m/s} \) is the initial speed, \( a_t = 2.5 \, \text{m/s}^2 \) is the tangential acceleration, and \( t = 2.0 \, \text{s} \). Substitute these values to find the final speed: \( v = 5.0 \, \text{m/s} + (2.5 \, \text{m/s}^2 \times 2.0 \, \text{s}) = 10.0 \, \text{m/s} \).
03

Calculate Centripetal Acceleration

The centripetal acceleration is given by the formula \( a_c = \frac{v^2}{r} \), where \( v = 10.0 \, \text{m/s} \) is the tangential speed after 2 seconds and \( r = 32 \, \text{m} \) is the radius of the circular turn. Substitute the values into the formula: \( a_c = \frac{(10.0 \, \text{m/s})^2}{32 \, \text{m}} = 3.125 \, \text{m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangential Acceleration
In circular motion, tangential acceleration refers to how fast an object's speed is changing along the path of the circle. It is directly related to the net tangential force acting on the object. According to Newton's second law, the relationship between force, mass, and acceleration is described with the formula:
  • Net force = mass × acceleration
For this speedboat example, with a given force of 550 N and a mass of 220 kg, the tangential acceleration is calculated by rearranging the formula:
  • \( a_t = \frac{F_t}{m} \)
By substituting the values:
  • \( a_t = \frac{550 \, \text{N}}{220 \, \text{kg}} = 2.5 \, \text{m/s}^2 \)
This indicates that the speed of the boat increases by 2.5 meters per second every second due to the engine's force during the circular turn. It's important to understand how tangential force and mass work together to result in this acceleration.
Centripetal Acceleration
Centripetal acceleration is a crucial concept in circular motion. It is the acceleration that keeps an object moving in a circular path, always directed toward the center of the circle. In our speedboat problem, after 2 seconds in the turn, the boat's speed increases from 5.0 m/s to 10.0 m/s. This new speed is used to calculate the centripetal acceleration.
The formula is:
  • \( a_c = \frac{v^2}{r} \)
Here, \( v \) is the tangential speed after 2 seconds, and \( r \) is the radius of the circular path (32 m). Substituting the known values:
  • \( a_c = \frac{(10.0 \, \text{m/s})^2}{32 \, \text{m}} = 3.125 \, \text{m/s}^2 \)
Remember, centripetal acceleration doesn't change the speed of the object, but changes the direction of the object's velocity, ensuring it continues along a circular path.
Newton's Second Law
Newton's second law of motion is a fundamental principle that links force, mass, and acceleration. In simple terms, it states:
  • A net force acting on a body produces an acceleration that is inversely proportional to its mass and directly proportional to the force.
This law is described by the equation:
  • \( F = ma \)
Where:
  • \( F \) is the force acting on the object,
  • \( m \) is the object's mass, and
  • \( a \) is the resulting acceleration.
In our problem, we specifically applied Newton's second law to determine tangential acceleration. With the net tangential force known, calculating the acceleration involved simply dividing this force by the boat's mass:
  • \( a_t = \frac{550 \, \text{N}}{220 \, \text{kg}} = 2.5 \, \text{m/s}^2 \)
This illustrates how changes in either force or mass can profoundly impact acceleration, highlighting an essential aspect of dynamics in circular motion. Newton's second law is pivotal for understanding the mechanics behind varied acceleration types, such as both tangential and centripetal seen in our problem.

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Most popular questions from this chapter

A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner \(A\) is \(n\) times as great as that measured at corner \(B\). What is the ratio \(L_{1} / L_{2}\) of the lengths of the sides of the rectangle when \(n=2.00 ?\)

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